A sentence S is given, composed of words separated by spaces. Each word consists of lowercase and uppercase letters only.

We would like to convert the sentence to " Goat Latin"  (a made-up language similar to Pig Latin.)

The rules of Goat Latin are as follows:

• If a word begins with a vowel (a, e, i, o, or u), append "ma" to the end of the word.
  For example, the word 'apple' becomes 'applema'.
   
• If a word begins with a consonant (i.e. not a vowel), remove the first letter and append it to the end, then add "ma".
  For example, the word "goat" becomes "oatgma".
   
• Add one letter 'a' to the end of each word per its word index in the sentence, starting with 1.
  For example, the first word gets "a" added to the end, the second word gets "aa" added to the end and so on.

Return the final sentence representing the conversion from S to Goat Latin. 

Example 1:

Input: "I speak Goat Latin"
Output: "Imaa peaksmaaa oatGmaaaa atinLmaaaaa"

Example 2:

Input: "The quick brown fox jumped over the lazy dog"
Output: "heTmaa uickqmaaa rownbmaaaa oxfmaaaaa umpedjmaaaaaa overmaaaaaaa hetmaaaaaaaa azylmaaaaaaaaa ogdmaaaaaaaaaa"

Notes:

• S contains only uppercase, lowercase and spaces. Exactly one space between each word.
• 1 <= S.length <= 150.

这道题让我们将一句话转为山羊拉丁文，有几个规则，如果单词是元音开头的，那么直接在但此后加ma，（为啥不是咩mie，西方的羊就是叫ma么，不知为何想起了老友记中Janice的笑声-.-|||，但其实Janice在剧中的声音是有意为之的，看过演员本人的访谈节目，相当正常，不得不说演员啊，声优啊，都是怪物，扯远了～），如果是非元音开头的单词，那么把首字母移到末尾，并且加ma。还有就是根据当前是第几个单词，后面加几个a，估计是为了模仿羊叫声，拉长音，感觉更像绵羊音一样。此题难度不是很大，就照题目要求来做，不需要啥特别的技巧。首先为了快速检测开头字母是否为元音，我们将所有元音加入一个HashSet，注意大小写的元音都要加进去。然后要一个单词一个单词的处理，这里我们使用C++的字符串流类来快速的提取单词，对于每个提取出的单词，我们先加上一个空格，然后判断开头字母是否为元音，是的话直接加上，不然就去子串去掉首字母，然后将首字母加到末尾。后面再加上ma，还有相对应数目个a。这里我们定义一个变量cnt，初始化为1，每处理一个单词，cnt自增1，这样我们就知道需要加的a的个数了，最后别忘了把结果res的首空格去掉，参见代码如下：

class Solution {
public:
    string toGoatLatin(string S) {
        unordered_set<char> vowel{'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
        istringstream ss(S);
        string res, t;
        int cnt = 1;
        while (ss >> t) {
            res += ' ' + (vowel.count(t[0]) ? t : t.substr(1) + t[0]) + "ma" + string(cnt, 'a');
            ++cnt;
        }
        return res.substr(1);
    }
};

参考资料：

https://leetcode.com/problems/goat-latin/

https://leetcode.com/problems/goat-latin/discuss/126973/Short-C%2B%2B-solution-using-io-stringstream

https://leetcode.com/problems/goat-latin/discuss/127024/C%2B%2BJavaPython-Straight-Forward-Solution

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